.TH std::remainder,std::remainderf,std::remainderl 3 "2024.06.10" "http://cppreference.com" "C++ Standard Libary"
.SH NAME
std::remainder,std::remainderf,std::remainderl \- std::remainder,std::remainderf,std::remainderl

.SH Synopsis
   Defined in header <cmath>
   float       remainder ( float x, float y );

   double      remainder ( double x, double y );                (until C++23)

   long double remainder ( long double x, long double y
   );
   constexpr /* floating-point-type */

               remainder ( /* floating-point-type */ x,         (since C++23)
                                                        \fB(1)\fP
                           /* floating-point-type */ y
   );
   float       remainderf( float x, float y );              \fB(2)\fP \fI(since C++11)\fP
                                                                (constexpr since C++23)
   long double remainderl( long double x, long double y     \fB(3)\fP \fI(since C++11)\fP
   );                                                           (constexpr since C++23)
   Additional overloads \fI(since C++11)\fP
   Defined in header <cmath>
   template< class Integer >                                (A) (constexpr since C++23)
   double      remainder ( Integer x, Integer y );

   1-3) Computes the IEEE remainder of the floating point division operation x / y.
   The library provides overloads of std::remainder for all cv-unqualified
   floating-point types as the type of the parameters.
   (since C++23)

   A) Additional overloads are provided for all integer types, which are  \fI(since C++11)\fP
   treated as double.

   The IEEE floating-point remainder of the division operation x / y calculated by this
   function is exactly the value x - quo * y, where the value quo is the integral value
   nearest the exact value x / y. When |quo - x / y| = ½, the value quo is chosen to be
   even.

   In contrast to std::fmod, the returned value is not guaranteed to have the same sign
   as x.

   If the returned value is zero, it will have the same sign as x.

.SH Parameters

   x, y - floating-point or integer values

.SH Return value

   If successful, returns the IEEE floating-point remainder of the division x / y as
   defined above.

   If a domain error occurs, an implementation-defined value is returned (NaN where
   supported).

   If a range error occurs due to underflow, the correct result is returned.

   If y is zero, but the domain error does not occur, zero is returned.

.SH Error handling

   Errors are reported as specified in math_errhandling.

   Domain error may occur if y is zero.

   If the implementation supports IEEE floating-point arithmetic (IEC 60559),

     * The current rounding mode has no effect.
     * FE_INEXACT is never raised, the result is always exact.
     * If x is ±∞ and y is not NaN, NaN is returned and FE_INVALID is raised.
     * If y is ±0 and x is not NaN, NaN is returned and FE_INVALID is raised.
     * If either argument is NaN, NaN is returned.

.SH Notes

   POSIX requires that a domain error occurs if x is infinite or y is zero.

   std::fmod, but not std::remainder is useful for doing silent wrapping of
   floating-point types to unsigned integer types: (0.0 <= (y = std::fmod(std::rint(x),
   65536.0)) ? y : 65536.0 + y) is in the range [-0.0, 65535.0], which corresponds to
   unsigned short, but std::remainder(std::rint(x), 65536.0) is in the range
   [-32767.0, +32768.0], which is outside of the range of signed short.

   The additional overloads are not required to be provided exactly as (A). They only
   need to be sufficient to ensure that for their first argument num1 and second
   argument num2:

     * If num1 or num2 has type long double, then std::remainder(num1,
       num2) has the same effect as std::remainder(static_cast<long
       double>(num1),
                      static_cast<long double>(num2)).
     * Otherwise, if num1 and/or num2 has type double or an integer type,
       then std::remainder(num1, num2) has the same effect as             (until C++23)
       std::remainder(static_cast<double>(num1),
                      static_cast<double>(num2)).
     * Otherwise, if num1 or num2 has type float, then
       std::remainder(num1, num2) has the same effect as
       std::remainder(static_cast<float>(num1),
                      static_cast<float>(num2)).
   If num1 and num2 have arithmetic types, then std::remainder(num1,
   num2) has the same effect as std::remainder(static_cast</*
   common-floating-point-type */>(num1),
                  static_cast</* common-floating-point-type */>(num2)),
   where /* common-floating-point-type */ is the floating-point type with
   the greatest floating-point conversion rank and greatest
   floating-point conversion subrank between the types of num1 and num2,  (since C++23)
   arguments of integer type are considered to have the same
   floating-point conversion rank as double.

   If no such floating-point type with the greatest rank and subrank
   exists, then overload resolution does not result in a usable candidate
   from the overloads provided.

.SH Example


// Run this code

 #include <cfenv>
 #include <cmath>
 #include <iostream>
 // #pragma STDC FENV_ACCESS ON

 int main()
 {
     std::cout << "remainder(+5.1, +3.0) = " << std::remainder(5.1, 3) << '\\n'
               << "remainder(-5.1, +3.0) = " << std::remainder(-5.1, 3) << '\\n'
               << "remainder(+5.1, -3.0) = " << std::remainder(5.1, -3) << '\\n'
               << "remainder(-5.1, -3.0) = " << std::remainder(-5.1, -3) << '\\n';

     // special values
     std::cout << "remainder(-0.0, 1.0) = " << std::remainder(-0.0, 1) << '\\n'
               << "remainder(5.1, Inf) = " << std::remainder(5.1, INFINITY) << '\\n';

     // error handling
     std::feclearexcept(FE_ALL_EXCEPT);
     std::cout << "remainder(+5.1, 0) = " << std::remainder(5.1, 0) << '\\n';
     if (fetestexcept(FE_INVALID))
         std::cout << "    FE_INVALID raised\\n";
 }

.SH Possible output:

 remainder(+5.1, +3.0) = -0.9
 remainder(-5.1, +3.0) = 0.9
 remainder(+5.1, -3.0) = -0.9
 remainder(-5.1, -3.0) = 0.9
 remainder(-0.0, 1.0) = -0
 remainder(5.1, Inf) = 5.1
 remainder(+5.1, 0) = -nan
     FE_INVALID raised

.SH See also

   div(int)
   ldiv     computes quotient and remainder of integer division
   lldiv    \fI(function)\fP
   \fI(C++11)\fP
   fmod
   fmodf    remainder of the floating point division operation
   fmodl    \fI(function)\fP
   \fI(C++11)\fP
   \fI(C++11)\fP
   remquo
   remquof
   remquol  signed remainder as well as the three last bits of the division operation
   \fI(C++11)\fP  \fI(function)\fP
   \fI(C++11)\fP
   \fI(C++11)\fP
   C documentation for
   remainder
